问题
Example question from a past operating system final, how do I calculate this kind of question?
A computer has a 64-bit virtual address space and 2048-byte pages. A page table entry takes 4 bytes. A multi-level page table is used because each table must be contained within a page. How many levels are required?
How would I calculate this?
回答1:
Since page table must fit in a page, page table size is 2048 bytes and each entry is 4 bytes thus a table holds 2048/4=512 entries. To address 512 entries it requires log2(512)=9 bits. The total number of bits available to encode the entry for each page level is 64-log2(2048)=53 bits (the number of bits of address space minus the page offset bits). Thus the total number of levels required is 53/9=6 (rounded up).
The x86-64 default page table size is 4096 bytes, each page table must fit in a page and a page table entry is 8 bytes. Current CPUs only implement 48 bits of virtual address space. How many page table levels are required?
回答2:
- Logical Address bit=64,
- Number of page will be= 2^64/2048 = 2^64/2^11 = 2^53
- Pages we have entry sine of page table= 4 Byte ,
- Number of Entry in 1 Page will be= 2048/4=>512,
- bit To represent one Entry=Log(512)=9bit,
- and bit for Page is= 53bit
- Therefore Number of Level =53/9=>6 Level Page Table
来源：https://stackoverflow.com/questions/5558886/multi-level-page-tables-hierarchical-paging